If \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are vectors such that \( |\vec{a}|=5,|\vec{b}|=12,|\vec{c}|=13 \) and \( \vec{a}+\vec{b}+\vec{c}=\vec{0} \) , then Match List-I with List-II
| LIST-I | LIST-II | ||
| A. | \( \vec{a}.\vec{c}+\vec{b}.\vec{c} \) | I. | −25 |
| B. | \( \vec{a}.\vec{b}+\vec{a}.\vec{c} \) | II. | −338 |
| C. | \( \vec{a}.\vec{b}+\vec{b}.\vec{c} \) | III. | −169 |
| D. | \( 2\left(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\right) \) | IV. | −144 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
Since \( \vec{a}+\vec{b}+\vec{c}=\vec{0} \), the sum of any two vectors is minus the third, e.g. \( \vec{a}+\vec{b}=-\vec{c} \).
(A) \( \vec{a}.\vec{c}+\vec{b}.\vec{c} \)
This matches III. −169.
\( \displaystyle (\vec{a}+\vec{b})\cdot\vec{c}=-\vec{c}\cdot\vec{c} \)
\( \displaystyle =-|\vec{c}|^2=-169 \)
(B) \( \vec{a}.\vec{b}+\vec{a}.\vec{c} \)
This matches I. −25.
\( \displaystyle \vec{a}\cdot(\vec{b}+\vec{c})=-\vec{a}\cdot\vec{a} \)
\( \displaystyle =-|\vec{a}|^2=-25 \)
(C) \( \vec{a}.\vec{b}+\vec{b}.\vec{c} \)
This matches IV. −144.
\( \displaystyle \vec{b}\cdot(\vec{a}+\vec{c})=-\vec{b}\cdot\vec{b} \)
\( \displaystyle =-|\vec{b}|^2=-144 \)
(D) \( 2\left(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\right) \)
This matches II. −338. Squaring \( \vec{a}+\vec{b}+\vec{c}=\vec{0} \):
\( \displaystyle |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0 \)
\( \displaystyle 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=-(25+144+169) \)
\( \displaystyle =-338 \)